Question

Lead (II) chloride (PbCl2) has a Ksp of 1.17 x 10-5. Calculate the molar solubility of PbCl2, in moles per liter. Enter your answer in scientific notation, in the form 1.23E-4. You do not need ot include units in your answer

Answer 1

Answer:

1.43 * 10^-2

Explanation:

Firstly, we need to get the ionization equation for lead I chloride. We can then get the Initial, change and equilibrium table of the lead and chloride ions that can help us calculate the molar solubility. This is shown as follows:

PbCl2(s) --> Pb2+(aq) + 2 Cl-(aq)

This shows that one mole of lead ii chloride will yield 1 mole of the lead ion and 2 moles of the chloride ion.

The ICE table is shown below:

Pb2+ 2Cl-

Initial 0.00 0.00

Change +s +2s

Equilibrium s 2s

Ksp = [Pb2+][Cl-]^2

Ksp = [s][2s]^2

Ksp = 4s^3

s^3 = Ksp/4

s = cube.root [Ksp/4]

s = cube.root[1.17 * 10^-5/4]

s = 1.43 * 10^-2